Start with a very ordinary-looking equation: x² + 1 = 0. Move the 1 across and you get x² = −1, so x = ±√(−1). Here the trouble begins. The square of any real number — positive or negative — is never negative, so no real number can be the square root of −1. For centuries, mathematicians simply left such roots alone. Then, slowly, they realised that giving this root a name opens a whole new number system that is genuinely useful in pure mathematics, applied mathematics, finance and computing.
That name is i, read as “iota”, and it is defined by i = √(−1), which means i² = −1. A complex number is any number of the form z = a + ib, where a and b are real numbers. The set of all complex numbers is denoted by C. So 3 + 4i, 2 − (5/7)i and −7 − 2i are all complex numbers. In z = a + ib, a is called the real part and b the imaginary part, written Re z = a and Im z = b. Notice a small trap: the imaginary part is the real number b itself, not ib. For z = 3 + 4i, Re z = 3 and Im z = 4.
📌 Book note: every real number is also a complex number, with 0 as its imaginary part. So 5 is really 5 + 0i. And the product of a non-zero real number with i is an imaginary number: 2i, −3i, √5 i and −(11/2)i are all imaginary numbers.
Worked idea: to split a number into its parts, just look at what stands alone and what stands with i. In −7 − 2i the real part is −7 and the imaginary part is −2 — the minus sign belongs to the part, so do not drop it.
مساوات x² + 1 = 0 کو حل کریں تو x² = −1 بنتا ہے، یعنی x = ±√(−1)۔ کسی بھی حقیقی عدد کا مربع منفی نہیں ہوتا، اس لیے حقیقی اعداد میں −1 کا جزر موجود نہیں۔ اسی مشکل سے مخلوط اعداد کا تصور پیدا ہوا۔
√(−1) کو i کہا جاتا ہے (آئیوٹا پڑھیں) اور i² = −1 ہوتا ہے۔ ہر مخلوط عدد z = a + ib کی صورت میں لکھا جاتا ہے، جہاں a اور b حقیقی اعداد ہیں۔ a کو حقیقی حصہ (Re z) اور b کو خیالی حصہ (Im z) کہتے ہیں۔ مثلاً z = 3 + 4i میں Re z = 3 اور Im z = 4 ہے۔ یاد رکھیں: خیالی حصہ صرف b ہے، ib نہیں۔
📌 کتابی نوٹ: ہر حقیقی عدد بھی ایک مخلوط عدد ہے، جس کا خیالی حصہ صفر ہوتا ہے، جیسے 5 = 5 + 0i۔
Once i exists, the ordinary rules of algebra keep working — with one extra fact always in your pocket: wherever you see i², replace it by −1. The book states the four basic operations for z = a + ib and w = c + id, where a, b, c, d and k are real numbers. Addition is like-with-like: (a + ib) + (c + id) = (a + c) + i(b + d). A real number times a complex number just scales both parts: k(a + ib) = ka + ikb. Subtraction is addition of the negative: (a + ib) − (c + id) = (a − c) + i(b − d). Multiplication needs the usual expansion, then the i² fix: (a + ib)(c + id) = ac + iad + ibc + i²bd = (ac − bd) + i(ad + bc).
Next comes a very handy partner for every complex number: the conjugate. If z = a + ib, its conjugate, written z̄ (z-bar), is a − ib — the same number with the sign of the imaginary part flipped. So the conjugate of 5 + 4i is 5 − 4i, and the conjugate of −2 + 3i is −2 − 3i. You will meet the conjugate again and again, because multiplying a number by its conjugate kills the imaginary part: (a + ib)(a − ib) = a² + b², a plain real number. That is exactly the trick used to divide complex numbers.
📌 Book note: a real number is self-conjugate. Flipping the sign of a zero imaginary part changes nothing, so for real a, ā = a.
Division in one line: to divide by c + id, multiply top and bottom by its conjugate c − id. The bottom becomes c² + d², a real number, and the top expands normally. Book Example 2 does exactly this for (4 + 2i) ÷ (3 − i) and lands on 1 + i — the full steps are worth copying once by hand, because the same move solves half this chapter.
مخلوط اعداد پر جمع، تفریق اور ضرب عام الجبرا کے اصولوں سے ہوتی ہے، بس یاد رکھیں کہ i² = −1 ہے۔ جمع میں حقیقی حصے حقیقی حصوں سے اور خیالی حصے خیالی حصوں سے ملتے ہیں۔ ضرب میں پہلے عام طریقے سے پھیلائیں، پھر i² کی جگہ −1 لکھ دیں۔
کانجوگیٹ (z̄) کسی عدد کا وہ ساتھی ہے جس میں خیالی حصے کی علامت بدل دی جاتی ہے: اگر z = a + ib ہو تو z̄ = a − ib۔ مثلاً 5 + 4i کا کانجوگیٹ 5 − 4i ہے۔ کسی عدد کو اس کے کانجوگیٹ سے ضرب دیں تو جواب حقیقی عدد a² + b² بنتا ہے، اور یہی ترکیب تقسیم میں کام آتی ہے۔
📌 کتابی نوٹ: حقیقی عدد خود اپنا کانجوگیٹ ہوتا ہے، کیونکہ اس کا خیالی حصہ صفر ہے۔
There is a second, tidier way to define the same numbers, and the book uses it to prove things cleanly. A complex number can be written as an ordered pair (a, b) of real numbers, so the set C is really R × R with special rules: (a, b) = (c, d) exactly when a = c and b = d; (a, b) + (c, d) = (a + c, b + d); (a, b)(c, d) = (ac − bd, ad + bc); and k(a, b) = (ka, kb). Subtraction follows at once: (a, b) − (c, d) = (a − c, b − d). The pair (a, b) and the form a + ib are the same number in two costumes — (3, 2) is 3 + 2i, and (−2, 3) is −2 + 3i.
Book Example 1 puts the pair rules to work on (8, 9) and (5, −6): sum (13, 3), difference (3, 15), product (94, −3). Check the product slowly — (8·5 − 9·(−6), 8·(−6) + 9·5) = (40 + 54, −48 + 45) = (94, −3). The minus signs are where students slip, so write each product out before adding.
The book then observes that C satisfies all the field axioms — it behaves like the real numbers under addition and multiplication. The landmarks: the additive identity is (0, 0); every (a, b) has additive inverse (−a, −b); the multiplicative identity is (1, 0); and every non-zero (a, b) has a multiplicative inverse, namely (a/(a² + b²), −b/(a² + b²)). That last formula is just “divide by the conjugate” in pair language, and Exercise 1.1 Q1 is built on it. Distributivity also holds: (a, b)[(c, d) ± (e, f)] = (a, b)(c, d) ± (a, b)(e, f).
📌 Book note: C does not satisfy the order axioms. There is no sensible way to say one complex number is “greater” or “less” than another, so never write z₁ < z₂ for complex numbers.
مخلوط عدد کو مرتب جوڑے (a, b) کی صورت میں بھی لکھا جا سکتا ہے۔ جوڑوں پر جمع، ضرب اور مساوات کے اصول وہی ہیں جو a + ib پر لاگو ہوتے ہیں، مثلاً (a, b)(c, d) = (ac − bd, ad + bc)۔ اس لیے (8, 9) اور (5, −6) کا حاصل ضرب (94, −3) بنتا ہے۔
مخلوط اعداد کا مجموعہ فیلڈ کے تمام اصول پورے کرتا ہے: جمعی شناخت (0, 0)، ضربی شناخت (1, 0)، اور ہر غیر صفر عدد کا ضربی معکوس (a/(a² + b²), −b/(a² + b²)) موجود ہے۔
📌 کتابی نوٹ: مخلوط اعداد پر ترتیب (بڑا/چھوٹا) کے اصول لاگو نہیں ہوتے، اس لیے دو مخلوط اعداد کا موازنہ < یا > سے نہیں کیا جاتا۔
Here the chapter turns visual, and it is worth slowing down for. Every complex number matches exactly one point of the coordinate plane, and every point matches exactly one complex number. The number x + iy is plotted as the point (x, y). In this picture the x-axis is renamed the real axis and the y-axis the imaginary axis; the whole plane is called the complex plane or the z-plane. A figure showing one or more complex numbers on this plane is an Argand diagram. Points on the real axis are real numbers, points on the imaginary axis are imaginary numbers, and everything else is a genuine mix.
The book's Figure (i) plots four numbers at once: A(3, 2) for 3 + 2i, B(−2, 2) for −2 + 2i, C(−3, −2) for −3 − 2i and D(2, −2) for 2 − 2i — one in each quadrant, which is a neat way to remember that the sign pattern of (x, y) decides the quadrant. Then comes the modulus: the real number |x + iy| = √(x² + y²), which is simply the distance from the origin to the point. In Figure (ii), if A is (x, y), then OA is that distance, by Pythagoras.
Book Example 3 ties the ideas together. For z = (1 + 2i)²/(2 − i), first simplify: (1 + 2i)² = −3 + 4i, then divide by (2 − i) using its conjugate (2 + i) to get z = −2 + i. Its conjugate is z̄ = −2 − i, and |z̄| = √((−2)² + (−1)²) = √5. Note the useful fact hiding here: |z̄| = |z| always, because squaring kills the sign of the imaginary part.
ہر مخلوط عدد x + iy کو مستوی میں نقطہ (x, y) سے ظاہر کیا جاتا ہے۔ اس مستوی میں x-محور کو حقیقی محور اور y-محور کو خیالی محور کہتے ہیں، اور پورے مستوی کو مخلوط مستوی یا z-مستوی کہتے ہیں۔ اس پر بنائی گئی شکل کو ارگنڈ ڈائیگرام کہتے ہیں۔
موڈولس |x + iy| = √(x² + y²) ہے، یعنی اصل نقطہ (origin) سے اس نقطے کا فاصلہ۔ مثلاً −2 + i کا کانجوگیٹ −2 − i ہے اور اس کا موڈولس √5 ہے۔ یاد رکھیں: |z̄| ہمیشہ |z| کے برابر ہوتا ہے۔
Two complex numbers are equal exactly when their real parts match and their imaginary parts match: a + bi = c + di ⟺ a = c and b = d. That one sentence is a problem-solving machine. Whenever an equation mixes x and y inside a complex number, split it into two real equations and solve them together. Book Example 4 does this with (3 + 2i)(x + iy) = 5 + 12i. Expanding gives (3x − 2y) + (2x + 3y)i = 5 + 12i, so 3x − 2y = 5 and 2x + 3y = 12. Multiply the first by 3 and the second by 2, add, and x = 3 falls out; putting x = 3 back gives y = 2.
Then the book derives a formula for the square root of a complex number. Suppose w = p + iq is a square root of z = x + iy. Squaring gives p² − q² = x and 2pq = y. A little algebra with (p² + q²)² = (p² − q²)² + 4p²q² leads to p² + q² = √(x² + y²) = |z|, and from there to p = ±√((|z| + x)/2) and q = ±√((|z| − x)/2). The signs of p and q are not free: since y = 2pq, they have the same sign when y > 0, opposite signs when y < 0, and one of them is zero when y = 0. The book packs all this into one formula: √z = ±( √((|z| + x)/2) + (iy/|y|)√((|z| − x)/2) ).
Book Example 5 applies it to 5 + 12i. Here |z| = 13, so √z = ±(√((13 + 5)/2) + i√((13 − 5)/2)) = ±(3 + 2i). The two square roots are 3 + 2i and −3 − 2i, plotted in the book as opposite points on an Argand diagram — as they must be, since one is the negative of the other.
دو مخلوط اعداد برابر تب ہوتے ہیں جب ان کے حقیقی حصے برابر اور خیالی حصے برابر ہوں۔ اس لیے کسی مساوات میں x اور y معلوم کرنے کے لیے اسے دو حقیقی مساواتوں میں توڑ لیں، پھر انہیں ایک ساتھ حل کریں۔
کسی مخلوط عدد z = x + iy کا جزر معلوم کرنے کا فارمولا یہ ہے: √z = ±( √((|z| + x)/2) + (iy/|y|)√((|z| − x)/2) )، جہاں |z| = √(x² + y²) ہے۔ y کی علامت سے p اور q کی علامتیں طے ہوتی ہیں: y > 0 ہو تو دونوں ایک جیسی علامت کے، y < 0 ہو تو مخالف علامت کے۔ مثلاً 5 + 12i کے جزر 3 + 2i اور −3 − 2i ہیں۔
A complex polynomial P(z) is a polynomial in the complex variable z with complex coefficients, in the usual form P(z) = aₙzⁿ + aₙ₋₁zⁿ⁻¹ + … + a₁z + a₀, where aₙ ≠ 0 and n ≥ 0 is the degree. If n = 0, P(z) is just a constant. The big fact of this section is the Fundamental Theorem of Algebra: a polynomial of degree n ≥ 1 has exactly n roots in C. Its corollary is the working tool: any such polynomial factors completely as P(z) = a(z − z₁)(z − z₂)…(z − zₙ), where the z's are the roots of P(z) = 0. That is why a quadratic with no real roots, like z² + 4, still factors over C as (z + 2i)(z − 2i).
📌 Book note: the values of z that satisfy P(z) = 0 are called the zeros of the function P(z) and the roots of the equation P(z) = 0 — two names for the same numbers.
The book's worked examples show three different ways in. Example 6 groups terms: z² + (i − 3)z − 3i = z(z + i) − 3(z + i) = (z + i)(z − 3). Example 7 rewrites a constant with i²: z² − 4iz + 12 becomes z² − 6iz + 2iz − i²12 = (z − 6i)(z + 2i). Example 8 takes out a common z first, then groups: z³ + (1 + i)z² + iz = z(z + 1)(z + i). And Example 9 brings in the Rational Root Theorem — for a polynomial with integer coefficients, any rational root p/q in lowest terms has p dividing the constant term and q dividing the leading coefficient — to spot that z = −1 is a root of z³ − 3z² + z + 5, then uses synthetic division and the quadratic formula to finish as (z + 1)(z − 2 − i)(z − 2 + i).
مخلوط کثیر رقمی P(z) وہ کثیر رقمی ہے جس کے ضابطے مخلوط اعداد ہوں۔ الجبرا کا بنیادی قضیہ کہتا ہے کہ درجہ n ≥ 1 کی کثیر رقمی کے مخلوط اعداد میں ٹھیک n جزر ہوتے ہیں، اور اسے خطی عوامل کے حاصل ضرب میں لکھا جا سکتا ہے: P(z) = a(z − z₁)(z − z₂)…(z − zₙ)۔
📌 کتابی نوٹ: P(z) = 0 کو پورا کرنے والی قدریں P(z) کے صفر (zeros) اور مساوات کے جزر (roots) کہلاتی ہیں۔
عوامل بنانے کے لیے کبھی مشترک عامل نکالیں، کبھی گروپ بندی کریں، اور کبھی عقلی جزر قضیہ (Rational Root Theorem) سے ممکنہ جزر آزمائیں، پھر مصنوعی تقسیم (synthetic division) سے آگے بڑھیں۔
Completing the square is the method you already know from earlier classes, and here it earns its keep: it solves every quadratic, even one that refuses to factor, and it hands you complex roots without any drama. The recipe never changes. First make the coefficient of z² equal to 1. Then move the constant to the other side. Then add the square of half the z-coefficient to both sides, so the left side becomes a perfect square. Take square roots — and here, for the first time, a negative number under the root is no obstacle, because √(−16) = 4i.
Book Example 10 works 2z² − 12z + 50 = 0 all the way through. Divide by 2: z² − 6z + 25 = 0, so z² − 2(3)z = −25. Add 3² to both sides: (z − 3)² = −16, so z − 3 = ±4i and z = 3 ± 4i. Then the corollary from the last section turns the roots straight into factors: 2z² − 12z + 50 = 2(z − 3 − 4i)(z − 3 + 4i). Notice the pattern in every answer of Exercise 1.3 Q4: complex roots of a real quadratic always arrive as a conjugate pair, a + bi and a − bi. If you ever get a lone complex root for a real quadratic, you have made an arithmetic slip — go back and check the signs.
مربع مکمل کرنے کا طریقہ ہر دو درجی مساوات کو حل کرتا ہے، چاہے وہ عوامل میں نہ ٹوٹتی ہو۔ پہلے z² کا ضابطہ 1 بنائیں، پھر مستقل رقم دوسری طرف لے جائیں، پھر z کے ضابطے کے نصف کا مربع دونوں طرف جمع کریں تاکہ بائیں طرف مکمل مربع بن جائے۔ پھر جزر لیں — منفی عدد کا جزر اب کوئی مسئلہ نہیں، کیونکہ √(−16) = 4i ہے۔
مثلاً 2z² − 12z + 50 = 0 سے (z − 3)² = −16 بنتا ہے، اس لیے z = 3 ± 4i۔ حقیقی ضابطوں والی مساوات کے مخلوط جزر ہمیشہ کانجوگیٹ جوڑے میں آتے ہیں۔
Now for the chapter's most famous little cast of characters. Let x be a cube root of unity, so x³ = 1 and x³ − 1 = 0. Factoring gives (x − 1)(x² + x + 1) = 0. The first factor gives x = 1, the ordinary root. The quadratic gives x = (−1 ± √(−3))/2 = (−1 ± √3 i)/2. So the three cube roots of unity are 1, (−1 + √3 i)/2 and (−1 − √3 i)/2.
📌 Book note: the two roots containing i are called the imaginary cube roots of unity.
The book names one of them ω (omega). If ω = (−1 + √3 i)/2, then the other one is ω², and if you swap the names, the statement still holds — each imaginary root is the square of the other. That gives the three properties the book lists, and they are worth memorising cold because Exercise 1.4 runs on them: (i) each complex cube root is the square of the other; (ii) the sum of all three roots is zero, 1 + ω + ω² = 0; (iii) the product of all three is unity, 1·ω·ω² = ω³ = 1, so each imaginary root is the reciprocal of the other, ω = 1/ω² and ω² = 1/ω. From (ii) you also get the handy rewrites ω² = −1 − ω and 1 + ω = −ω², which make Exercise 1.4 Q5 and Q6 fall apart nicely.
Book Example 11 is a sweet application: x³ + y³ = (x + y)(x + ωy)(x + ω²y). Multiply the last two factors, use ω + ω² = −1 and ω³ = 1, and the middle collapses to x² − xy + y², giving the familiar factorisation with a complex twist.
اگر x وحدت کا مکعب جزر ہو تو x³ = 1، یعنی (x − 1)(x² + x + 1) = 0۔ اس سے تین جزر ملتے ہیں: 1، (−1 + √3 i)/2 اور (−1 − √3 i)/2۔ i والے دونوں جزر کو خیالی مکعب جزر کہتے ہیں۔
📌 کتابی نوٹ: i پر مشتمل جزر خیالی مکعب جزر کہلاتے ہیں۔
ایک خیالی جزر کو ω (اومیگا) کہیں تو دوسرا ω² ہوتا ہے۔ تین اہم خصوصیات یاد رکھیں: 1 + ω + ω² = 0، ω³ = 1، اور ہر خیالی جزر دوسرے کا معکوس ہے۔ انہی سے مشق ۱.۴ کے زیادہ تر سوالات حل ہوتے ہیں۔
The fourth roots follow the same idea with even less work. Let x be a fourth root of unity: x⁴ = 1, so x⁴ − 1 = 0, which factors as (x² − 1)(x² + 1) = 0. From x² = 1 come x = ±1, and from x² = −1 come x = ±i. Hence the four fourth roots of unity are 1, −1, i, −i — a neat little cross on the Argand diagram.
Their properties mirror the cube case: (i) the sum of all four is zero, since 1 + (−1) + i + (−i) = 0; (ii) the real roots 1 and −1 are additive inverses of each other; (iii) the imaginary roots i and −i are conjugates of each other; (iv) the product of all four is −1, since 1 × (−1) × i × (−i) = −1. Watch the last one: i × (−i) = −i² = 1, and then 1 × (−1) × 1 = −1. Students often rush this and write 1, so do it in two bites as shown.
Exercise 1.4 Q2 generalises the pattern: the fourth roots of 16 are ±2 and ±2i, of 81 are ±3 and ±3i, and of 625 are ±5 and ±5i — in each case just scale the unit cross by the real fourth root, and the sum is still zero.
اگر x وحدت کا چہارم جزر ہو تو x⁴ = 1، یعنی (x² − 1)(x² + 1) = 0۔ اس سے چار جزر ملتے ہیں: 1، −1، i اور −i۔ ان کا مجموعہ صفر ہے، حقیقی جزر ایک دوسرے کے جمعی معکوس ہیں، خیالی جزر ایک دوسرے کے کانجوگیٹ ہیں، اور چاروں کا حاصل ضرب −1 ہے۔
اسی اصول سے 16 کے چہارم جزر ±2 اور ±2i، اور 81 کے ±3 اور ±3i بنتے ہیں — بس اکائی والے جزر کو حقیقی چہارم جزر سے ضرب دے دیں۔
So far every point has been named by (x, y): how far across, how far up. Polar coordinates name the same point differently — by how far and in which direction. A point P is written (r, θ), where r is the directed distance from a fixed origin O, called the pole, and θ is the angle the line OP makes with the polar axis, which is just the positive x-axis. The book's polar grid picture, with its circles and spokes at 0°, 45°, 90° and so on, is worth studying: each circle is a fixed r, each spoke a fixed θ.
Usually r is taken non-negative (r ≥ 0), but the book also allows r < 0, and the meaning is precise: when r > 0, θ is measured from the polar axis to OP in the usual way, so (5, π/4) is the point 5 units out at π/4. When r < 0, the point sits |r| units in the opposite direction, at angle θ + π. So (−5, π/4) is 5 units out at 5π/4 — the same point as (5, 5π/4).
📌 Book note: (−5, π/4) and (5, 5π/4) represent the same point in the plane. More generally, polar coordinates are not unique: adding 2π to θ, or flipping the sign of r and adding π to θ, leaves the point unchanged. That non-uniqueness is exactly what Exercise 1.5 Q1 is testing when it asks you to plot points with negative r and angles beyond one full turn.
قطبی نظام میں نقطے کو (x, y) کی بجائے (r, θ) سے ظاہر کیا جاتا ہے: r اصل نقطے (pole) سے فاصلہ ہے اور θ قطبی محور (مثبت x-محور) سے بننے والا زاویہ ہے۔ عام طور پر r ≥ 0 لیا جاتا ہے، لیکن r منفی بھی ہو سکتا ہے۔
اگر r > 0 ہو تو نقطہ θ کی سمت میں r فاصلے پر ہوتا ہے، جیسے (5, π/4)۔ اگر r < 0 ہو تو نقطہ مخالف سمت میں |r| فاصلے پر ہوتا ہے، یعنی زاویہ θ + π پر؛ اس لیے (−5, π/4) اور (5, 5π/4) ایک ہی نقطہ ہیں۔
📌 کتابی نوٹ: (−5, π/4) اور (5, 5π/4) مستوی میں ایک ہی نقطے کو ظاہر کرتے ہیں۔
Put the Argand picture and the polar picture together and you get the polar form. For z = x + iy, the book's triangle gives x = r cos θ and y = r sin θ, where r = |z| = √(x² + y²) is the modulus and θ is called an argument of z. Hence x + iy = r(cos θ + i sin θ), with θ = tan⁻¹(y/x) when x ≠ 0. Book Example 12 converts 1 + i√3: r² = 1 + 3 = 4 so r = 2, and θ = tan⁻¹(√3) = 60°, giving 2(cos 60° + i sin 60°).
Because θ can be increased by 2π without moving the point, the book singles out one value, the principal argument, written Arg z, chosen in the range (−π, π]. The book's note handles the axes: if x = 0 and y > 0 then θ = 90°; if x = 0 and y < 0 then θ = −90°; if x = 0 and y = 0 then θ is undefined; if y = 0 and x > 0 then θ = 0°; if y = 0 and x < 0 then θ = 180°. In general, use the quadrant, not just the tan⁻¹ button — a point in the third quadrant and a point in the first can share the same y/x ratio.
Polar form then makes multiplication and division almost trivial. For z₁ = r₁(cos θ₁ + i sin θ₁) and z₂ = r₂(cos θ₂ + i sin θ₂): multiply the moduli and add the arguments, z₁z₂ = r₁r₂[cos(θ₁ + θ₂) + i sin(θ₁ + θ₂)]; divide the moduli and subtract the arguments, z₁/z₂ = (r₁/r₂)[cos(θ₁ − θ₂) + i sin(θ₁ − θ₂)]. Book Example 13 multiplies 5(cos π/6 + i sin π/6) by 4(cos 3π/2 + i sin 3π/2) to get 20(cos 5π/3 + i sin 5π/3), and Example 14 divides (2/7)(cos 7π/6 + i sin 7π/6) by (3/5)(cos(−π/2) + i sin(−π/2)) to get (10/21)(cos 5π/3 + i sin 5π/3). The book numbers this operations section 1.3.3 even though it sits inside the polar material — a numbering quirk, flagged here so it does not confuse you.
⚠️ Honest note (derived, not a book passage): the book writes Arg(z₁z₂) = Arg z₁ + Arg z₂. For general arguments that is exact, but for the principal argument it holds only up to adding or subtracting 2π to bring the sum back into (−π, π]. Exercise 1.5 Q6 is solved in that sense.
قطبی صورت میں z = x + iy کو z = r(cos θ + i sin θ) لکھا جاتا ہے، جہاں r = |z| = √(x² + y²) موڈولس ہے اور θ دلیل (argument) ہے۔ مثلاً 1 + i√3 کی قطبی صورت 2(cos 60° + i sin 60°) ہے۔
بنیادی دلیل (Arg z) وہ خاص دلیل ہے جو (−π, π] کے درمیان لی جاتی ہے۔ محوروں پر واقع نقطوں کے لیے کتاب نے الگ اصول دیے ہیں، جیسے x = 0، y > 0 ہو تو θ = 90°۔
قطبی صورت میں ضرب آسان ہو جاتی ہے: موڈولس ضرب اور دلیلیں جمع؛ تقسیم میں موڈولس تقسیم اور دلیلیں تفریق۔ یعنی arg(z₁z₂) = arg(z₁) + arg(z₂) اور arg(z₁/z₂) = arg(z₁) − arg(z₂)۔
The chapter closes by showing that none of this is decoration. In an alternating-current circuit, resistance generalises to impedance Z. If R is resistance, XL inductive reactance and XC capacitive reactance, the net reactance is X = XL − XC, and impedance is the complex number Z = R + iX. Ohm's law for AC circuits then reads V = I·Z, with voltage, current and impedance all in complex (polar) form. Book Example 17 divides a voltage of 25(cos 30° + i sin 30°) V by an impedance of 11(cos 55.35° + i sin 55.35°) Ω in polar form — moduli divide, arguments subtract — to get a current of about 2.04 − 0.95i A. Examples 15 and 16, just before this section, do the same kind of geometric work with moduli and arguments: a modulus equation can describe a straight line in the complex plane, and an argument condition can describe a circle.
The second application is cryptography: turning readable plaintext into secret ciphertext with a key, and back again. Book Example 18 encrypts the word “MATH” by assigning A = 1, B = 2, …, Z = 26, writing each letter as a complex number with zero imaginary part, and multiplying by the key k = 2 + 3i. M = 13 becomes (13)(2 + 3i) = 26 + 39i, and so on. Decryption divides by k — which is the same as multiplying by its multiplicative inverse — and every letter returns exactly. Exercise 1.5 Q20–Q22 ring the changes on this: a different multiplying key, and finally a key that is added instead of multiplied, so decryption subtracts it.
اے سی سرکٹ میں مزاحمت کی جگہ امپیڈنس Z استعمال ہوتی ہے: Z = R + iX، جہاں R مزاحمت اور X = XL − XC ری ایکٹنس ہے۔ اوہم کا قانون اس صورت میں V = I·Z بن جاتا ہے۔ قطبی صورت میں تقسیم سے کرنٹ آسانی سے نکل آتا ہے۔
دوسری عملی مثال خفیہ نگاری (cryptography) ہے: ہر حرف کو عدد دیا جاتا ہے (A = 1، …، Z = 26)، پھر اسے کلید (key) سے ضرب دے کر خفیہ پیغام بنایا جاتا ہے، اور کلید کے معکوس سے تقسیم کر کے اصل لفظ واپس لایا جاتا ہے۔
Below are all questions of Exercises 1.1, 1.2, 1.3, 1.4 and 1.5 from Unit 1 of the official book (printed pages 5, 7–8, 11, 13, 20–21), each solved step by step. The book has no multiple-choice section, no review exercise and no inquisitive/numerical section in this unit, so none is invented here. Where a step is derived rather than quoted from a book passage, the answer says so.
Counting rule used here: one numbered question = one answer block; its parts (i), (ii), … are solved inside that block. That gives 9 + 12 + 8 + 9 + 22 = 60 blocks, covering 131 parts in total.
Method: the inverse of (a, b) is (a/(a² + b²), −b/(a² + b²)). (i) a² + b² = 16 + 49 = 65, so the inverse is (−4/65, −7/65), i.e. −4/65 − (7/65)i. Check: (−4 + 7i)(−4/65 − 7i/65) = (16 + 49)/65 = 1. (ii) a² + b² = 2 + 5 = 7, so the inverse is (√2/7, √5/7), i.e. √2/7 + (√5/7)i. (iii) a² + b² = 1, so the inverse is (1, 0) itself — as expected, since 1 is the multiplicative identity.
(i) Multiply top and bottom by the conjugate 4 − 5i: (2 − 7i)(4 − 5i)/(16 + 25) = (8 − 10i − 28i + 35i²)/41 = (−27 − 38i)/41. So the real part is −27/41 and the imaginary part is −38/41. (ii) (−2 + 3i)² = 4 − 12i + 9i² = −5 − 12i. Divide by (1 + i) using (1 − i): (−5 − 12i)(1 − i)/2 = (−5 + 5i − 12i + 12i²)/2 = (−17 − 7i)/2. (iii) i(1 − i)/2 = (i − i²)/2 = (1 + i)/2. (iv) (4 + 3i)² = 16 + 24i + 9i² = 7 + 24i. Divide by (4 − 3i) using (4 + 3i): (7 + 24i)(4 + 3i)/25 = (28 + 21i + 96i + 72i²)/25 = (−44 + 117i)/25.
Write z = a + ib with a, b real, so z̄ = a − ib. (⇒) If z̄ = z, then a − ib = a + ib, so comparing imaginary parts, −b = b, giving 2b = 0, i.e. b = 0. Hence z = a, a real number. (⇐) If z is real, then b = 0, so z = a and z̄ = a = z. Both directions hold, so z̄ = z if and only if z is real.
Write z = a + ib, z̄ = a − ib. (i) z + z̄ = (a + ib) + (a − ib) = 2a, so (z + z̄)/2 = a = Re(z). (ii) z − z̄ = (a + ib) − (a − ib) = 2ib, so (z − z̄)/(2i) = b = Im(z). (iii) z·z̄ = (a + ib)(a − ib) = a² − i²b² = a² + b², and |z|² = (√(a² + b²))² = a² + b². Hence |z|² = z·z̄.
First z₁z₃ = (2 + i)(1 + 3i) = 2 + 6i + i + 3i² = −1 + 7i. Its conjugate is −1 − 7i. Now divide by z₂ = 3 − 2i using its conjugate 3 + 2i: (−1 − 7i)(3 + 2i)/(9 + 4) = (−3 − 2i − 21i − 14i²)/13 = (11 − 23i)/13. So the answer is (11/13) − (23/13)i.
(i) 2z₁ − 4z₂ = (4 + 14i) − (−20 + 12i) = 24 + 2i. Its modulus is √(24² + 2²) = √580 = 2√145. (ii) z̄₁ = 2 − 7i, so 3z₁ + 2z̄₁ = (6 + 21i) + (4 − 14i) = 10 + 7i. Modulus = √(100 + 49) = √149. (iii) −7z₂ + 2z̄₂ = (35 − 21i) + (−10 − 6i) = 25 − 27i. Modulus = √(625 + 729) = √1354. (iv) z₁ + z₂ = −3 + 10i, whose modulus is √109. The modulus of a cube is the cube of the modulus, so |(z₁ + z₂)³| = (√109)³ = 109√109.
Take iⁿ common: the sum equals iⁿ(i + i² + i³ + i⁴). Now i² = −1, i³ = −i and i⁴ = 1, so the bracket is i + (−1) + (−i) + 1 = 0. Anything times 0 is 0, so the whole sum is 0 for every natural number n.
First simplify the base: (1 + i)/(1 − i) = (1 + i)(1 + i)/(1 + 1) = (1 + 2i + i²)/2 = 2i/2 = i. So the equation becomes (i)²ⁿ = 1, i.e. (i²)ⁿ = (−1)ⁿ = 1. This holds exactly when n is even. The least positive even n is n = 2. Check: ((1 + i)/(1 − i))⁴ = i⁴ = 1.
Divide n by 4: n = 4q + r with 0 ≤ r < 4. Then iⁿ = i⁴ᑫ⁺ʳ = (i⁴)ᑫ · iʳ. Since i⁴ = 1, (i⁴)ᑫ = 1, leaving iⁿ = iʳ. (This argument works for every n, not only n > 4; the book states it for n > 4.)
(i) Right side: (5 − i)(3 + 4i) = 15 + 20i − 3i − 4i² = 19 + 17i; times i gives −17 + 19i. Left side is (x + 2) + i(y − 3). Equating parts: x + 2 = −17 gives x = −19; y − 3 = 19 gives y = 22. (ii) Right side: (2 − 3i)(−5 + 5i) = 5 + 25i, and −i³/5 = i/5 (since i³ = −i), so the product is (5 + 25i)(i/5) = −5 + i. Left side: (x + iy)(1 − i) = (x + y) + i(y − x). So x + y = −5 and y − x = 1. Adding gives 2y = −4, y = −2, and then x = −3. (iii) Write each fraction with its conjugate: x(2 − i)/5 + y(3 + i)/10 = 4 + 5i. Multiply by 10: (4x + 3y) + i(−2x + y) = 40 + 50i. So −2x + y = 50 and 4x + 3y = 40. From the first, y = 50 + 2x; substituting gives 4x + 150 + 6x = 40, so x = −11 and y = 28.
Rearrange: z₂ = z₁ − (−27 + 15i) = (−13 + 24i) + 27 − 15i = 14 + 9i. Comparing with z₂ = x + yi gives x = 14 and y = 9. Check: (−13 + 24i) − (14 + 9i) = −27 + 15i. ✓
Method for all three: (x + iy)² = (x² − y²) + i(2xy). Equate parts, then use x² + y² = √((x² − y²)² + (2xy)²) to get x² and y² separately; the sign of xy fixes whether x and y share a sign. (i) x² − y² = 25, 2xy = 60. Then x² + y² = √(625 + 3600) = 65, so x² = 45 and y² = 20, with xy > 0. Hence (x, y) = (3√5, 2√5) or (−3√5, −2√5). (ii) x² − y² = 64, 2xy = 48. Then x² + y² = √(4096 + 2304) = 80, so x² = 72 and y² = 8, with xy > 0. Hence (x, y) = (6√2, 2√2) or (−6√2, −2√2). (iii) First simplify the right side: (2i − 3)(3 − i)/10 = (−7 + 9i)/10. So x² − y² = −7/10 and 2xy = 9/10. Then x² + y² = √(49/100 + 81/100) = √130/10. Hence x² = (√130 − 7)/20 and y² = (√130 + 7)/20, with xy > 0, so x and y take the same sign: x = ±√((√130 − 7)/20), y = ±√((√130 + 7)/20), both + or both −.
As printed, z₂ = 1 − α is a real number (no i is shown after α). Then z₁z₂ = (2 + 3i)(1 − α) = 2(1 − α) + 3(1 − α)i, so Im(z₁z₂) = 3(1 − α). Setting this equal to 7 gives 1 − α = 7/3, so α = −4/3. Check: 3(1 − (−4/3)) = 3 × 7/3 = 7. ✓
Add the two equations: 2z₁ = 16 + 6i, so z₁ = 8 + 3i, giving x = 8 and y = 3. Subtract the second from the first: 2z₂ = 4 + 2i, so z₂ = 2 + i, giving a = 2 and b = 1. Check: (8 + 3i) + (2 + i) = 10 + 4i and (8 + 3i) − (2 + i) = 6 + 2i. ✓
Write z₁ = a + ib and z₂ = c + id. Then z₁z₂ = (ac − bd) + i(ad + bc), so its conjugate is (ac − bd) − i(ad + bc). On the other side, z̄₁z̄₂ = (a − ib)(c − id) = (ac − bd) + i(−ad − bc) = (ac − bd) − i(ad + bc). Both sides agree, so the conjugate of a product is the product of the conjugates.
Use √z = ±( √((|z| + x)/2) + (iy/|y|)√((|z| − x)/2) ). (i) |z| = 25: ±(√9 − i√16) = ±(3 − 4i). Check: (3 − 4i)² = 9 − 24i − 16 = −7 − 24i. ✓ (ii) |z| = 10: ±(√9 − i√1) = ±(3 − i). (iii) |z| = 39: ±(√12 − i√27) = ±(2√3 − 3√3 i). Check: (2√3)² − (3√3)² = 12 − 27 = −15 and 2(2√3)(−3√3) = −36. ✓ (iv) |z| = 169: ±(√144 + i√25) = ±(12 + 5i).
Here x = 13, y = −20√3 and |z| = √(169 + 1200) = √1369 = 37. By the square-root formula: ±( √((37 + 13)/2) − i√((37 − 13)/2) ) = ±(5 − 2√3 i), since y < 0 makes the imaginary part negative. So the two square roots are 5 − 2√3 i and −5 + 2√3 i. On an Argand diagram they are the points (5, −2√3) in the fourth quadrant and (−5, 2√3) in the second quadrant, opposite each other through the origin.
Right side: i(11 − i) = 11i − i² = 1 + 11i. Left side: (−7 + i)(x + iy) = (−7x − y) + i(x − 7y); adding (−1 − 5i) gives (−7x − y − 1) + i(x − 7y − 5). Equate parts: −7x − y − 1 = 1, i.e. 7x + y = −2; and x − 7y − 5 = 11, i.e. x − 7y = 16. From the first, y = −2 − 7x; substituting gives x + 14 + 49x = 16, so 50x = 2, x = 1/25, and y = −2 − 7/25 = −57/25.
Right side: i(11 − i) − 4i = 11i + 1 − 4i = 1 + 7i. Left side: (5 − 2i)(x + iy) + 3 = (5x + 2y + 3) + i(−2x + 5y). Equate parts: 5x + 2y + 3 = 1, i.e. 5x + 2y = −2; and −2x + 5y = 7. Multiply the first by 5 and the second by 2: 25x + 10y = −10 and −4x + 10y = 14. Subtracting gives 29x = −24, so x = −24/29. Then 5y = 7 + 2x = 7 − 48/29 = 155/29, so y = 31/29.
Clear denominators with conjugates: (u − 2)(2 − i)/5 + (v − 3)(2 + i)/5 = 4i. Multiply by 5 and expand: the real part is 2(u − 2) + 2(v − 3) = 2u + 2v − 10, and the imaginary part is −(u − 2) + (v − 3) = −u + v − 1. Equating with 0 + 20i: 2u + 2v − 10 = 0, i.e. u + v = 5; and −u + v − 1 = 20, i.e. −u + v = 21. Adding: 2v = 26, v = 13; then u = 5 − 13 = −8. Check: (−10)/(2 + i) + 10/(2 − i) = −2(2 − i) + 2(2 + i) = 4i. ✓
z₁z₂ = (4 + 5i)(α − 2i) = 4α − 8i + 5αi − 10i² = (4α + 10) + i(5α − 8). So Re(z₁z₂) = 4α + 10. Setting this equal to 20 gives 4α = 10, so α = 5/2.
Key idea: a sum of squares factors over C with i: p² + q² = (p + qi)(p − qi). (i) (a + 2bi)(a − 2bi). (ii) (3a + 4bi)(3a − 4bi). (iii) 3(x² + y²) = 3(x + iy)(x − iy). (iv) (12x + 15yi)(12x − 15yi). (v) Notice (z − i)² = z² − 2iz + i² = z² − 2iz − 1, so this is (z − i)² (a repeated factor; the root i has multiplicity 2). (vi) Roots from the quadratic formula: z = (−6 ± √(36 − 52))/2 = −3 ± 2i, so (z + 3 − 2i)(z + 3 + 2i). (vii) Roots −2 ± i, so (z + 2 − i)(z + 2 + i). (viii) Discriminant 484 − 520 = −36, so z = (22 ± 6i)/4 = (11 ± 3i)/2, giving 2(z − (11 + 3i)/2)(z − (11 − 3i)/2).
(i) z³ + 8 = (z + 2)(z² − 2z + 4); the quadratic has roots 1 ± √3 i, so (z + 2)(z − 1 − √3 i)(z − 1 + √3 i). (ii) z³ + 27 = (z + 3)(z² − 3z + 9); roots of the quadratic are (3 ± 3√3 i)/2, so (z + 3)(z − (3 + 3√3 i)/2)(z − (3 − 3√3 i)/2). (iii) Group: z²(z − 2) + 16(z − 2) = (z − 2)(z² + 16) = (z − 2)(z + 4i)(z − 4i). (iv) Put w = z²: w² + 21w − 100 = 0 gives w = (−21 ± 29)/2, i.e. w = 4 or w = −25. So (z² − 4)(z² + 25) = (z − 2)(z + 2)(z + 5i)(z − 5i). (v) (z² − 4)(z² + 4) = (z − 2)(z + 2)(z + 2i)(z − 2i). (vi) w² + 3w − 4 = (w + 4)(w − 1), so (z² + 4)(z² − 1) = (z + 2i)(z − 2i)(z − 1)(z + 1). (vii) (z² + 2)(z² + 3) = (z + √2 i)(z − √2 i)(z + √3 i)(z − √3 i). (viii) w² − 32w − 3969 = 0 gives w = (32 ± 130)/2, i.e. w = 81 or w = −49. So (z² − 81)(z² + 49) = (z − 9)(z + 9)(z + 7i)(z − 7i).
Put w = z²: w² + 7w − 144 = (w + 16)(w − 9) = 0, so w = −16 or w = 9. From z² = 9 come z = 3 and z = −3; from z² = −16 come z = 4i and z = −4i. Hence the roots are 3, −3, 4i, −4i, and z⁴ + 7z² − 144 = (z − 3)(z + 3)(z − 4i)(z + 4i).
(i) Divide by 2: z² − (3/2)z + 2 = 0. Add (3/4)² both sides: (z − 3/4)² = 9/16 − 2 = −23/16, so z = 3/4 ± (√23/4)i. (ii) (z − 3)² = 9 − 30 = −21, so z = 3 ± √21 i. (iii) Divide by 3: z² − 6z + 50/3 = 0. Add 9: (z − 3)² = 9 − 50/3 = −23/3, so z = 3 ± i√(23/3) = 3 ± (√69/3)i. (iv) (z + 2)² = 4 − 13 = −9, so z = −2 ± 3i. (v) Divide by 2: z² + 3z + 9/2 = 0. Add (3/2)²: (z + 3/2)² = 9/4 − 9/2 = −9/4, so z = −3/2 ± (3/2)i. (vi) Divide by 3: z² − (5/3)z + 7/3 = 0. Add (5/6)²: (z − 5/6)² = 25/36 − 84/36 = −59/36, so z = 5/6 ± (√59/6)i.
(i) z⁴ = 16, so z² = ±4, giving z = 2, −2, 2i, −2i. (ii) z(z⁴ − 81) = 0. So z = 0 or z⁴ = 81, giving z² = ±9, i.e. z = 3, −3, 3i, −3i. All together: 0, 3, −3, 3i, −3i. (iii) z(z⁴ − 1) = 0. So z = 0 or z⁴ = 1, giving z² = ±1, i.e. z = 1, −1, i, −i. All together: 0, 1, −1, i, −i. (iv) Group: z²(z − 5) + (z − 5) = (z − 5)(z² + 1) = 0, so z = 5, i, −i. (v) Put w = z²: 4w² − 25w − 21 = 0 gives w = (25 ± 31)/8, i.e. w = 7 or w = −3/4. So z = √7, −√7, (√3/2)i, −(√3/2)i. (vi) Group: z²(z + 1) + (z + 1) = (z + 1)(z² + 1) = 0, so z = −1, i, −i.
With leading coefficient a, P(z) = a(z − 3)(z + 2i)(z − 2i) = a(z − 3)(z² + 4). Put z = 1: P(1) = a(−2)(5) = −10a. Setting −10a = 20 gives a = −2. So P(z) = −2(z − 3)(z² + 4) = −2z³ + 6z² − 8z + 24. Check: P(1) = −2 + 6 − 8 + 24 = 20. ✓
P(z) = a(z − 2i)(z + 2i)(z − 1)(z + 1) = a(z² + 4)(z² − 1). Put z = 2: P(2) = a(8)(3) = 24a. Setting 24a = 240 gives a = 10. So P(z) = 10(z² + 4)(z² − 1) = 10z⁴ + 30z² − 40.
P(z) = a(z − 4)(z + 4)(z − 1 − i)(z − 1 + i) = a(z² − 16)(z² − 2z + 2). Put z = 2: P(2) = a(−12)(2) = −24a. Setting −24a = 72 gives a = −3. So P(z) = −3(z² − 16)(z² − 2z + 2) = −3z⁴ + 6z³ + 42z² − 96z + 96. Check: P(2) = −48 + 48 + 168 − 192 + 96 = 72. ✓
Method: if r is the real cube root, the three roots are r, rω and rω², where ω = (−1 + √3 i)/2. (i) 2, 2ω = −1 + √3 i, 2ω² = −1 − √3 i. (ii) −2, −2ω = 1 − √3 i, −2ω² = 1 + √3 i. (iii) −3, −3ω = (3 − 3√3 i)/2, −3ω² = (3 + 3√3 i)/2. (iv) 4, 4ω = −2 + 2√3 i, 4ω² = −2 − 2√3 i. (v) −5, −5ω = (5 − 5√3 i)/2, −5ω² = (5 + 5√3 i)/2. In each case, cubing any listed root returns the original number, since ω³ = 1.
Method: scale the fourth roots of unity 1, −1, i, −i by the real fourth root. 16: ±2 and ±2i; sum 2 + (−2) + 2i + (−2i) = 0. 81: ±3 and ±3i; sum 3 + (−3) + 3i + (−3i) = 0. 625: ±5 and ±5i; sum 5 + (−5) + 5i + (−5i) = 0. Check one: (2i)⁴ = 16i⁴ = 16. ✓
If n is a multiple of 3, write n = 3m. Then ωⁿ = (ω³)ᵐ = 1ᵐ = 1, and ω²ⁿ = (ω³)²ᵐ = 1. So 1 + ωⁿ + ω²ⁿ = 1 + 1 + 1 = 3. Derived extra (not printed in the question, for completeness): if n is not a multiple of 3, then ωⁿ is ω or ω², and the sum is 1 + ω + ω² = 0. The book's wording here is compressed (“respectively”); the multiple-of-3 case above is what it asks.
(i) Note (−1 + √−3)/2 = (−1 + √3 i)/2 = ω, and the second term is ω². Since ω³ = 1, ω⁷ = ω⁶·ω = ω and (ω²)⁷ = ω¹⁴ = ω¹²·ω² = ω². Their sum is ω + ω² = −1 (from 1 + ω + ω² = 0). (ii) Here −1 + √−3 = −1 + √3 i = 2ω, and −1 − √−3 = 2ω². So the sum is (2ω)⁵ + (2ω²)⁵ = 32ω⁵ + 32ω¹⁰. Now ω⁵ = ω² and ω¹⁰ = ω, so the sum is 32(ω² + ω) = 32(−1) = −32.
Reduce every power mod 3 using ω³ = 1: ω⁴ = ω, ω⁸ = ω², ω¹⁶ = ω, and so on. So the factors alternate between A = 1 − ω + ω² and B = 1 − ω² + ω. Using ω² = −1 − ω: A = 1 − ω + (−1 − ω) = −2ω. Similarly B = 1 − ω² + ω = −2ω² (since ω = −1 − ω²). Hence each adjacent pair multiplies to AB = (−2ω)(−2ω²) = 4ω³ = 4. With 2n factors there are n such pairs, so the product is 4ⁿ = 2²ⁿ.
Square first: ((i + √3)/2)² = (i² + 2√3 i + 3)/4 = (2 + 2√3 i)/4 = (1 + √3 i)/2 = −ω² (since ω² = (−1 − √3 i)/2). Raising to the 4th power: ((i + √3)/2)⁸ = (−ω²)⁴ = ω⁸ = ω². Similarly ((i − √3)/2)² = (1 − √3 i)/2 = −ω, so ((i − √3)/2)⁸ = (−ω)⁴ = ω⁴ = ω. Adding: ω² + ω = −1, from 1 + ω + ω² = 0.
Write the six terms out: k = 0 gives 1; k = 1 gives ω²; k = 2 gives ω⁴ = ω; k = 3 gives ω⁶ = 1; k = 4 gives ω⁸ = ω²; k = 5 gives ω¹⁰ = ω. So the sum is 2(1 + ω + ω²) = 2 × 0 = 0.
Multiply the denominator by ω: ω(aω² + bω + c) = aω³ + bω² + cω = a + bω² + cω (using ω³ = 1), which is exactly the numerator. So the numerator equals ω times the denominator, and dividing gives ω — provided the denominator is not zero, which is the usual standing assumption for such an identity.
Reduce all powers mod 3: ω¹² = 1, ω¹⁷ = ω², ω¹⁹ = ω, so the numerator is a + bω² + cω. And ω¹⁴ = ω², ω²² = ω, ω³⁰ = 1, so the denominator is aω² + bω + c. By Q8, the ratio of exactly these two expressions is ω. (Read the last denominator power carefully as ω³⁰ — it reduces to 1, which is what makes the denominator match Q8.)
Remember: a negative r means “go |r| in the opposite direction”, i.e. add π to the angle. (i) 2 units at 75°, first quadrant. (ii) 3 units at 120° + 180° = 300°, fourth quadrant. (iii) 2 units at 30°, first quadrant. (iv) 5 units at 150°, second quadrant. (v) 5/2 units at π/3 + π = 4π/3, third quadrant. (vi) −2π/3 is −120°; opposite direction is 60°, so 3 units at 60°, first quadrant. (vii) −19π/12 = −285° ≡ 75°; opposite direction is 75° + 180° = 255°, so 9/2 units at 255°, third quadrant. (viii) Opposite of 5π/12 is 17π/12, so 5/2 units at 17π/12, third quadrant.
(i) r = 5, cos θ = 4/5, sin θ = 3/5, so 5(cos θ + i sin θ) with θ = tan⁻¹(3/4) ≈ 36.87°. (ii) r = √2, θ = 45°: √2(cos 45° + i sin 45°). (iii) r = √(1/4 + 3/4) = 1, θ = 60°: cos 60° + i sin 60°. (iv) r = 5, third quadrant, θ = −120° (equivalently 240°): 5(cos(−120°) + i sin(−120°)). (v) (1 − i)/(1 + i) = (1 − i)²/2 = −i, so r = 1, θ = −90°: cos(−90°) + i sin(−90°). (vi) (√3 + i)(1 − √3 i)/4 = (√3 − i)/2, so r = 1, θ = −30°: cos(−30°) + i sin(−30°). (vii) (3 + 4i)(4 − 3i)/25 = (24 + 7i)/25, so r = 1, θ = tan⁻¹(7/24) ≈ 16.26°: cos θ + i sin θ with cos θ = 24/25, sin θ = 7/25.
(i) cos 5π/3 = 1/2, sin 5π/3 = −√3/2, so z = 2 − 2√3 i. (ii) cos 7π/6 = −√3/2, sin 7π/6 = −1/2, so z = −(3√3/4) − (3/4)i. (iii) 23π/12 ≡ −π/12, i.e. −15°: cos = (√6 + √2)/4, sin = −(√6 − √2)/4, so z = (7(√6 + √2)/4) − i(7(√6 − √2)/4) ≈ 6.76 − 1.81i. (iv) cos(−11π/12) = −(√6 + √2)/4, sin(−11π/12) = −(√6 − √2)/4, so z = −(11(√6 + √2)/4) − i(11(√6 − √2)/4) ≈ −10.63 − 2.85i. (v) −17π/12 ≡ 7π/12, i.e. 105°: cos = (√2 − √6)/4, sin = (√6 + √2)/4, so z = (10/3)[(√2 − √6)/4 + i(√6 + √2)/4] ≈ −0.86 + 3.22i. (vi) No degree sign is printed; taking −33 as −33° (as in the degree examples nearby): z = 2 cos 33° − i 2 sin 33° ≈ 1.68 − 1.09i. ⚠️ If −33 were meant in radians, the value would differ — the book does not print a unit here, so degrees is the stated reading.
First convert: z₁ = 9(−√2/2 − i√2/2) = −(9√2/2)(1 + i) ≈ −6.36 − 6.36i; z₂ = 5(1/2 + i√3/2) = 2.5 + 4.33i. (i) z₁ + z₂ = (−9√2/2 + 5/2) + i(−9√2/2 + 5√3/2) ≈ −3.86 − 2.03i. (ii) z₁ − z₂ = (−9√2/2 − 5/2) + i(−9√2/2 − 5√3/2) ≈ −8.86 − 10.69i. (iii) Polar: r = 45, θ = 5π/4 + π/3 = 19π/12, so 45(cos 19π/12 + i sin 19π/12) = 45[(√6 − √2)/4 − i(√6 + √2)/4] ≈ 11.65 − 43.47i. (iv) Polar: r = 9/5, θ = 5π/4 − π/3 = 11π/12, so (9/5)(cos 11π/12 + i sin 11π/12) ≈ −1.74 + 0.47i.
Convert: 23π/12 ≡ −π/12, so z₁ = 7[(√6 + √2)/4 − i(√6 − √2)/4] ≈ 6.76 − 1.81i; z₂ = 11[−(√6 + √2)/4 + i(√6 − √2)/4] ≈ −10.63 + 2.85i. (i) Sum = −(√6 + √2) + i(√6 − √2) ≈ −3.86 + 1.04i. (ii) Difference = (9/2)(√6 + √2) − i(9/2)(√6 − √2) ≈ 17.39 − 4.66i. (iii) Polar: r = 77, θ = 23π/12 + 11π/12 = 17π/6 ≡ 5π/6, so 77(cos 5π/6 + i sin 5π/6) = −(77√3/2) + (77/2)i ≈ −66.68 + 38.50i. (iv) Polar: r = 7/11, θ = 23π/12 − 11π/12 = π, so (7/11)(cos π + i sin π) = −7/11.
Write z₁ = r₁(cos θ₁ + i sin θ₁) and z₂ = r₂(cos θ₂ + i sin θ₂), where θ₁, θ₂ are arguments. By the polar multiplication rule, z₁z₂ = r₁r₂[cos(θ₁ + θ₂) + i sin(θ₁ + θ₂)], so an argument of z₁z₂ is θ₁ + θ₂; similarly z₁/z₂ has argument θ₁ − θ₂. ⚠️ Range note (derived): for the principal argument Arg ∈ (−π, π], the sum θ₁ + θ₂ may fall outside that range, in which case the principal value is θ₁ + θ₂ ± 2π. The book's equalities are exact for general arguments and for principal arguments whenever the sum or difference already lies in (−π, π].
Divide moduli, subtract arguments: quotient = 2(cos(150° − 30°) + i sin(150° − 30°)) = 2(cos 120° + i sin 120°) = 2(−1/2 + i√3/2) = −1 + √3 i.
Multiply moduli, add arguments: product = 10(cos(60° + 90°) + i sin(60° + 90°)) = 10(cos 150° + i sin 150°) = 10(−√3/2 + i/2) = −5√3 + 5i.
Modulus: |z| = √((−2)² + (−2)²) = √8 = 2√2. The point (−2, −2) lies in the third quadrant, where the reference angle is tan⁻¹(2/2) = 45°, so the principal argument is −180° + 45° = −135° (equivalently 225°, or −3π/4 in radians).
z̄ − 2 + i = (x − 2) + i(1 − y). An argument of 2π/3 (second quadrant) means tan of the argument is (1 − y)/(x − 2) = tan(2π/3) = −√3, with x − 2 < 0 and 1 − y > 0. So 1 − y = −√3(x − 2), i.e. y = √3 x − 2√3 + 1, with x < 2 (and y > 1). ⚠️ Derived condition: the solution is a half-line (ray), not the whole line — the inequalities are part of the answer, because arg fixes a direction, not just a slope.
Note 9π/4 = 2π + π/4, so the argument is in the π/4 direction: the quotient has tan = 1. Now z̄ − 1 + 2i = (x − 1) + i(2 − y) and z̄ + 1 − 2i = (x + 1) + i(−y − 2). Multiplying the numerator by the conjugate of the denominator, the real part of the numerator is (x − 1)(x + 1) + (2 − y)(−y − 2) = (x² − 1) + (y² − 4) = x² + y² − 5, and the imaginary part is (2 − y)(x + 1) − (x − 1)(−y − 2) = 4x − 2y. Since tan = (imaginary part)/(real part) = tan(π/4) = 1, we get 4x − 2y = x² + y² − 5, i.e. x² + y² − 4x + 2y − 5 = 0, as required.
The difference of arguments is the argument of the quotient (z − 2 − 3i)/(z + 2 + 3i). A value of 2π means the quotient is a positive real number (argument 0 modulo 2π), so its imaginary part must be zero. Now (z − 2 − 3i) = (x − 2) + i(y − 3) and (z + 2 + 3i) = (x + 2) + i(y + 3). The imaginary part of their quotient's numerator (after multiplying by the conjugate) is (y − 3)(x + 2) − (x − 2)(y + 3) = 4y − 6x. Setting it to zero gives 4y = 6x, i.e. 2y = 3x. ⚠️ Note (derived): principal arguments lie in (−π, π], so their difference cannot literally equal 2π; the book's 2π is read as a general-argument value, i.e. the quotient's argument is 0 modulo 2π.
Square both sides. Left: |x + i(y − 2)|² = x² + (y − 2)². Right: |(x + 2) − iy|² = (x + 2)² + y². Equating: x² + y² − 4y + 4 = x² + 4x + 4 + y². Cancelling x², y² and 4 gives −4y = 4x, i.e. y = −x — the straight line through the origin at −45°.
Left: 5z + 4 + i = (5x + 4) + i(5y + 1). Right: 5z̄ − 3 + 2i = (5x − 3) + i(2 − 5y). Square both moduli: (5x + 4)² + (5y + 1)² = (5x − 3)² + (2 − 5y)². Expanding: 25x² + 40x + 16 + 25y² + 10y + 1 = 25x² − 30x + 9 + 4 − 20y + 25y². Cancelling the squares gives 40x + 10y + 17 = −30x − 20y + 13, i.e. 70x + 30y = −4, or 35x + 15y = −2 — a straight line.
Left: 3z̄ − 2 + i = (3x − 2) + i(1 − 3y), squared modulus (3x − 2)² + (1 − 3y)². Right: 3z + i = 3x + i(3y + 1), squared modulus 9x² + (3y + 1)². Equating and expanding: 9x² − 12x + 4 + 1 − 6y + 9y² = 9x² + 9y² + 6y + 1. Cancelling gives −12x − 12y + 4 = 0, i.e. x + y = 1/3. The solution set is that straight line in the complex plane.
Compute the squared moduli (z ≠ i, so the denominator is non-zero). 1 − iz = 1 − i(x + iy) = (1 + y) − ix, so |1 − iz|² = (1 + y)² + x². And |z − i|² = x² + (y − 1)². If |w| = 1, these are equal: (1 + y)² + x² = x² + (y − 1)². Expanding: 1 + 2y + y² = y² − 2y + 1, so 4y = 0, i.e. y = 0. Hence z = x, a real number.
Square the modulus and use |w|² = w·w̄. The numerator's squared modulus is (z₂ − z₁)(z̄₂ − z̄₁) = |z₂|² − z₂z̄₁ − z₁z̄₂ + |z₁|² = 1 − z₂z̄₁ − z₁z̄₂ + |z₁|², since |z₂|² = 1. The denominator's squared modulus is (1 − z̄₁z₂)(1 − z₁z̄₂) = 1 − z₁z̄₂ − z̄₁z₂ + |z₁|²|z₂|² = 1 − z₁z̄₂ − z̄₁z₂ + |z₁|². The two are identical (and the denominator is non-zero because z₁ ≠ z₂ with |z₂| = 1), so the squared modulus is 1, and the modulus itself is 1.
First write Z in polar form: |Z| = √(1/4 + 3/4) = 1 and arg Z = tan⁻¹(√3) = π/3, so Z = cos π/3 + i sin π/3. By Ohm's law for AC, I = V/Z. Divide moduli and subtract arguments: I = 120(cos(π/4 − π/3) + i sin(π/4 − π/3)) = 120(cos(−π/12) + i sin(−π/12)) A. In rectangular terms this is about 115.91 − 31.06i A.
Rectangular: I = V/Z = (90 + 30i)/(3 − 6i). Multiply top and bottom by 3 + 6i: (90 + 30i)(3 + 6i)/45 = (270 + 540i + 90i + 180i²)/45 = (90 + 630i)/45 = 2 + 14i A. Polar: |I| = √(4 + 196) = √200 = 10√2, and arg I = tan⁻¹(14/2) = tan⁻¹(7) ≈ 81.87°, first quadrant. So I = 10√2(cos(tan⁻¹7) + i sin(tan⁻¹7)) A, with cos = √2/10 and sin = 7√2/10.
Letters as numbers (A = 1, …, Z = 26): C = 3, O = 15, D = 4, E = 5. Encrypt (multiply by 2 − i): 3(2 − i) = 6 − 3i; 15(2 − i) = 30 − 15i; 4(2 − i) = 8 − 4i; 5(2 − i) = 10 − 5i. Decrypt: the inverse of 2 − i is (2 + i)/5. Multiply each ciphertext by it: (6 − 3i)(2 + i)/5 = (12 + 6i − 6i − 3i²)/5 = 15/5 = 3 → C; (30 − 15i)(2 + i)/5 = (60 + 30i − 30i − 15i²)/5 = 75/5 = 15 → O; (8 − 4i)(2 + i)/5 = (16 + 8i − 8i − 4i²)/5 = 20/5 = 4 → D; (10 − 5i)(2 + i)/5 = (20 + 10i − 10i − 5i²)/5 = 25/5 = 5 → E. The word “CODE” is recovered.
Letters: Q = 17, U = 21, I = 9, Z = 26. Encrypt (multiply by 3 − 3i): 17(3 − 3i) = 51 − 51i; 21(3 − 3i) = 63 − 63i; 9(3 − 3i) = 27 − 27i; 26(3 − 3i) = 78 − 78i. Decrypt: the inverse of 3 − 3i is (3 + 3i)/18 = (1 + i)/6. Multiply: (51 − 51i)(1 + i)/6 = (51 + 51i − 51i − 51i²)/6 = 102/6 = 17 → Q; (63 − 63i)(1 + i)/6 = 126/6 = 21 → U; (27 − 27i)(1 + i)/6 = 54/6 = 9 → I; (78 − 78i)(1 + i)/6 = 156/6 = 26 → Z. The word “QUIZ” is recovered.
Letters: C = 3, L = 12, A = 1, S = 19, S = 19. Encrypt (add −3 + 4i): 3 + (−3 + 4i) = 4i; 12 + (−3 + 4i) = 9 + 4i; 1 + (−3 + 4i) = −2 + 4i; 19 + (−3 + 4i) = 16 + 4i; and the second S likewise 16 + 4i. Decrypt (subtract the key): 4i − (−3 + 4i) = 3 → C; (9 + 4i) − (−3 + 4i) = 12 → L; (−2 + 4i) − (−3 + 4i) = 1 → A; (16 + 4i) − (−3 + 4i) = 19 → S; and again 19 → S. The word “CLASS” is recovered.
اس باب میں کیا ہے؟ Complex numbers کی پوری بنیاد: i = √(−1) سے شروع کر کے real/imaginary parts, conjugate, ordered pairs, Argand diagram اور modulus, برابری سے square root کا formula, polynomials کے linear factors, completing تھے square, cube اور fourth roots of unity (ω ki properties samet), polar coordinates, polar form اور polar operations, اور آخر میں impedance اور cryptography جیسے اصل استعمال.
Revision guidance: پہلے conjugate سے تقسیم اور modulus پکے کرو, پھر ω کی تین properties زبانی یاد کرو, پھر polar form کے چار اصول (zarb/taqseem) لکھ کر یاد کرو. Exercise 1.5 کے Q10–Q15 میں صرف modulus/argument کی شرطیں لکھنا سیکھنا ہے — پوری لمبی حل کی ضرورت نہیں ہوتی, شرط لکھو اور simplify کرو. Exercise: 📝 Official Exercise Questions — with Answers (60 questions, 131 parts).
اس باب میں مخلوط اعداد کی مکمل بنیاد ہے: i کی تعریف، حقیقی و خیالی حصے، کانجوگیٹ، ارگنڈ ڈائیگرام، موڈولس، جزر المربع کا فارمولا، کثیر رقمیوں کے عوامل، مربع مکمل کرنے کا طریقہ، وحدت کے مکعب و چہارم جزر، قطبی صورت اور حقیقی دنیا میں استعمال۔
امتحان کے لیے تین چیزیں سب سے اہم ہیں: کانجوگیٹ سے تقسیم، ω کی خصوصیات (1 + ω + ω² = 0، ω³ = 1)، اور قطبی صورت میں ضرب و تقسیم کے اصول۔ ان تینوں پر عبور حاصل کر لیں تو اس باب کی مشقیں آسان لگیں گی۔
Chapter-wise tests, fresh random papers اور instant score کے لیے PunjabTest.com کا میں test section دیکھو:
PunjabTest.com — Complete Guide Book • FSc Part 1 Mathematics, Chapter 1 (New Book, Session 2026–27) • English + Urdu